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Vladimir [108]
2 years ago
10

Suppose we have a 12.2 L sample containing 0.50 mol oxygen gas at a pressure of 1 atm and a temperature of 25 degrees celcius. I

f all this oxygen were converted to ozone at the same temperature and pressure, what would be the volume of the ozone?Balanced equation: 3O2 → 2O3a. 6.5Lb. 2.8Lc. 8.1Ld. 7.4L
Chemistry
1 answer:
snow_lady [41]2 years ago
6 0

Answer:

c. 8.1 L

Explanation:

Given that:-

Moles of oxygen gas = 0.50 mol

According to the reaction shown below as:-

3O_2\rightarrow 2O_3

3 moles of oxygen gas on reaction gives 2 moles of ozone

Also,

1 mole of oxygen gas on reaction gives 2/3 moles of ozone

So,

0.50 mole of oxygen gas on reaction gives \frac{2}{3}\times 0.50 moles of ozone

Moles of ozone = 0.3333 mol

Pressure = 1 atm

Temperature = 25.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25.0 + 273.15) K = 298.15 K  

Volume = ?

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V = 0.3333 mol × 0.0821 L.atm/K.mol × 298.15 K  

⇒V = 8.1 L

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What will be the end product for each electrode after electrolysis if the solution is concentrated aqueous sodium chloride ?​
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Answer:

Cathode: Hydrogen gas

Anode: Chlorine gas

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Anions: Cl⁻ , OH⁻

② Decide which ions are preferentially discharged.

These are the factors:

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For the discharge of anions,

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In this case:

For cations, H⁺ ions are selectively discharged at the cathode as its position is lower than Na⁺ in the reactivity series.

Anion: Cl⁻ ions, being more concentrated, are selectively discharged at the anode.

☆For electrolysis,

Cation at the cathode (-ve terminal)

Anion at the anode (+ve terminal)

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